Two wires of same material and same extension `Deltal` have lengths l and 3l and diameters 3d and d. What will be the ratio of the forces applied on the two?

Options

A. `1//27`
B. 27
C. 9
D. 3

Answer

To solve the problem, we need to find the ratio of the forces applied on two wires of the same material and same extension. The specifications of the wires are as follows:

- Wire 1: Length = \( l \), Diameter = \( 3d \)
- Wire 2: Length = \( 3l \), Diameter = \( d \)
- Both wires have the same extension \( \Delta l \).

### Step-by-Step Solution:

1. **Understand the relationship between stress, strain, and Young's modulus**:
- The stress \( \sigma \) in a wire is given by:
\[ \sigma = \frac{F}{A} \]
where \( F \) is the force applied and \( A \) is the cross-sectional area.
- The strain \( \epsilon \) is given by:
\[ \epsilon = \frac{\Delta l}{L} \]
where \( \Delta l \) is the extension and \( L \) is the original length.
- Young's modulus \( Y \) is defined as:
\[ Y = \frac{\sigma}{\epsilon} \]

2. **Calculate the cross-sectional areas**:
- For Wire 1 (diameter = \( 3d \)):
\[ A_1 = \frac{\pi (3d)^2}{4} = \frac{9\pi d^2}{4} \]
- For Wire 2 (diameter = \( d \)):
\[ A_2 = \frac{\pi d^2}{4} \]

3. **Express the stress and strain for both wires**:
- For Wire 1:
\[ \sigma_1 = \frac{F_1}{A_1} = \frac{F_1}{\frac{9\pi d^2}{4}} = \frac{4F_1}{9\pi d^2} \]
\[ \epsilon_1 = \frac{\Delta l}{l} \]
- For Wire 2:
\[ \sigma_2 = \frac{F_2}{A_2} = \frac{F_2}{\frac{\pi d^2}{4}} = \frac{4F_2}{\pi d^2} \]
\[ \epsilon_2 = \frac{\Delta l}{3l} \]

4. **Set up the equations using Young's modulus**:
- Since both wires are made of the same material, they have the same Young's modulus \( Y \):
\[ Y = \frac{\sigma_1}{\epsilon_1} = \frac{\sigma_2}{\epsilon_2} \]

5. **Substituting the expressions**:
- For Wire 1:
\[ Y = \frac{\frac{4F_1}{9\pi d^2}}{\frac{\Delta l}{l}} \implies Y = \frac{4F_1 l}{9\pi d^2 \Delta l} \]
- For Wire 2:
\[ Y = \frac{\frac{4F_2}{\pi d^2}}{\frac{\Delta l}{3l}} \implies Y = \frac{12F_2 l}{\pi d^2 \Delta l} \]

6. **Equating the two expressions for Young's modulus**:
\[ \frac{4F_1 l}{9\pi d^2 \Delta l} = \frac{12F_2 l}{\pi d^2 \Delta l} \]

7. **Cancel common terms**:
- Cancel \( l \), \( \pi \), and \( \Delta l \) from both sides:
\[ \frac{4F_1}{9} = 12F_2 \]

8. **Rearranging to find the ratio**:
\[ F_1 = 12 \cdot \frac{9}{4} F_2 \implies F_1 = 27 F_2 \]

9. **Final ratio of forces**:
\[ \frac{F_1}{F_2} = 27:1 \]

### Conclusion:
The ratio of the forces applied on the two wires is \( 27:1 \).